Q. Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located over a tower at one point and moved with velocities v_{1} = 3 m/s and v_{2} = 4 m/s horizontally in opposite directions. The time when their velocity vectors become mutually perpendicular is

(a) $ \displaystyle t= \sqrt{0.12}$ s

(b) $ \displaystyle t= \sqrt{0.14}$ s

(c) t = 1.2 s

(d)t = 0.5 s

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Sol: $ \displaystyle \vec{v_1′} = v_1\hat{i} + gt\hat{(-j)}$

$ \displaystyle \vec{v_2′} = v_2\hat{(-i)} + gt\hat{(-j)}$

When velocity vectors become mutually perpendicular

$ \displaystyle \vec{v_1′}.\vec{v_2′} = 0 $

$ \displaystyle (v_1\hat{i} + gt\hat{(-j)}).(v_2\hat{(-i)} + gt\hat{(-j)}) = 0$

$\displaystyle -v_1.v_2 + g^2 t^2 = 0 $

$\displaystyle t^2 = \frac{v_1 . v_2}{g^2}$

$ \displaystyle t = \sqrt{\frac{v_1 . v_2}{g^2}}$

$ \displaystyle = \sqrt{\frac{3\times 4}{100}} $

$ \displaystyle = \sqrt{0.12} sec$